# Homework Help: Calculations in a group

1. Oct 5, 2008

### SNOOTCHIEBOOCHEE

1. The problem statement, all variables and given/known data

assume that the equation xyz=1 holds in a group G. does it follow that yzx=1? yxz=1?

3. The attempt at a solution

My thoughts are the following:

Since in the problem, it does not state that G is an abelian group, we cannot assume the law of composition is commutitive. thus, it doesn not follow that yzx or yxz =1.

Is this correct?

2. Oct 5, 2008

### morphism

Can you come with examples to substantiate your claims?

3. Oct 5, 2008

### SNOOTCHIEBOOCHEE

Sure.

For example, lets say x,y,z were matrices that some how mutilpied to the identity matrix.

And since we all know that matrix multiplication is not commutitive, any other combination (rather than xyz) may not give us a result of the identity.

Last edited: Oct 5, 2008
4. Oct 5, 2008

### morphism

What if in this particular instance the matrices x, y and z did commute with each other? And when you're multiplying matrices, what group are you doing this in?

5. Oct 5, 2008

### SNOOTCHIEBOOCHEE

If i am multiplying matrices i am in the group G= (real numbers, matrix multiplication).

if, for a particualr case x y and z did commute then i wouldnt know what to do.

6. Oct 5, 2008

### morphism

Is that G really a group?

What I'm really asking you is this: can you explicitly give me a group G and elements x,y,z in G such that xyz=1 but yzx and yxz are not equal to 1? This will be enough to answer the questions you first post.

7. Oct 5, 2008

### Dick

But yzx IS equal to 1. If xyz=1 then yz=x^(-1). It's going to be hard to find a counterexample for that one.

8. Oct 5, 2008

### SNOOTCHIEBOOCHEE

hmm i guess that G would not be a group... would

G=(Square invertible Matrices of some constant size with real elements, matrix multiplication) work?

And if i could find some 3 matrices that work like that (which im sure i could w/ a little elbow greece) this would be enough justification?

(also can you tell me if i even got the right answer to the problem so i could stop wasting time if it wasnt right?)

Edit:

Dick, does this mean that yzx would always necessarily = one if xyz=1?

9. Oct 5, 2008

### morphism

Yes...

I don't think it was a waste of time. I was hoping you'd realize that your approach to the problem ("G is not abelian, so this can't happen!") was flawed.

10. Oct 5, 2008

### SNOOTCHIEBOOCHEE

Ok, but does this mean that yzx is always equal to 1 too? (i remember there being a property of group so that everything has to have an inverse).

and does yxz NOT equal 1? i cant see similar maniuplations leading to a result of one on the right side.